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If M, N are integers greater than 1 such that 2M<N, which of the foll
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18 Feb 2019, 09:10
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GMATH practice exercise (Quant Class 17) If M, N are integers greater than 1 such that 2M<N, which of the following numbers could be twice the value of the sum of all integers from M to N, including both of them? I. 54 II. 52 III. 50 (A) I. only (B) II. only (C) III. only (D) Exactly two of them (E) None of them The tricky alternative choice would be "All of them". We believe the examiner would NOT put it among the answer choices!
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Re: If M, N are integers greater than 1 such that 2M<N, which of the foll
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18 Feb 2019, 09:29
fskilnik wrote: GMATH practice exercise (Quant Class 17)
If M, N are integers greater than 1 such that 2M<N, which of the following numbers could be twice the value of the sum of all integers from M to N, including both of them?
I. 54 II. 52 III. 50
(A) I. only (B) II. only (C) III. only (D) Exactly two of them (E) None of them
Would still keep this, though brute force is a time consuming approach. 2M<N, N >2M, one thing to observe was that, the series will always start from an even Integer Will start from a single digit, 8,9,10,11,12 => 50 Now we can think greater values now 12,13,14,15 => 54 D
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If M, N are integers greater than 1 such that 2M<N, which of the foll
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18 Feb 2019, 18:25
fskilnik wrote: GMATH practice exercise (Quant Class 17)
If M, N are integers greater than 1 such that 2M<N, which of the following numbers could be twice the value of the sum of all integers from M to N, including both of them?
I. 54 II. 52 III. 50
(A) I. only (B) II. only (C) III. only (D) Exactly two of them (E) None of them 54, 52 and 50 are options for TWICE the sum. Thus, I, II, and III imply the following options for the ACTUAL SUM: I: 27 II: 26 III: 25 Case 1: M=2 Since it is required that N>2M, N≥5 M=2 and N=5 > sum = 2+3+4+5 = 14 M=2 and N=6 > sum increases by 6 > 14+6 = 20 M=2 and N=7 > sum increases by 7 > 20+7 = 27The green case indicates that M=2 and N=7 will yield a sum of 27, implying that option I  54  can be the value of TWICE the sum. Case 2: M=3 Since it is required that N>2M, N≥7 M=3 and N=7 > sum = 3+4+5+6+7 = 25The green case indicates that M=3 and N=7 will yield a sum of 25, implying that option III  50  can be the value of TWICE the sum. Thus, I and III are possible.
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Re: If M, N are integers greater than 1 such that 2M<N, which of the foll
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19 Feb 2019, 00:04
Hi, this is my first answer. Kindly check and suggest if my approach is right. Given, 2M<N => Assuming the AP has only 2 numbers in the series, M and N, M+N>3M. Thus solving 3M with the options (54/2, 52/2, 50/2) should give a reminder, suggesting that the sum of M+N is greater than the values. That is, 3M=27, No remainder. Therefore, M+N not greater than 3M. 3M=26, Remainder = 2. Therefore, M+N>3M. 3M=25, Remainder = 1. Therefore, M+N>3M. Hence, (ii) and (iii) could be the answers, hence D.
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Re: If M, N are integers greater than 1 such that 2M<N, which of the foll
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19 Feb 2019, 07:13
OhsostudiousMJ wrote: Hi, this is my first answer. Kindly check and suggest if my approach is right.
Given, 2M<N => Assuming the AP has only 2 numbers in the series, M and N, M+N>3M. Thus solving 3M with the options (54/2, 52/2, 50/2) should give a reminder, suggesting that the sum of M+N is greater than the values.
That is, 3M=27, No remainder. Therefore, M+N not greater than 3M. 3M=26, Remainder = 2. Therefore, M+N>3M. 3M=25, Remainder = 1. Therefore, M+N>3M.
Hence, (ii) and (iii) could be the answers, hence D.
Hi OhsostudiousMJ , Welcome to our GMAT Club community! You have discussed whether M+N is greater than 3M, but this is not our focus. This problem is hard, even in terms of understanding what is going on... Suggestion: start dealing with easier questions (600700 or below 600). You can find each question´s level looking at the questions tags when they are posted. Regards and success in your studies! Fabio.
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If M, N are integers greater than 1 such that 2M<N, which of the foll
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19 Feb 2019, 07:40
fskilnik wrote: GMATH practice exercise (Quant Class 17)
If M, N are integers greater than 1 such that 2M<N, which of the following numbers could be twice the value of the sum of all integers from M to N, including both of them?
I. 54 II. 52 III. 50
(A) I. only (B) II. only (C) III. only (D) Exactly two of them (E) None of them
Although the problem seems hard, the focusednumbers (half the numbers given) are SMALL... and that´s the hint to try the "organized manual work" (chosen in previous solutions)! We are looking for the 25 (III), 26 (II) and 27 (I) possibilities. \({S_K} = 1 + 2 + \ldots + K = {{K\left( {K + 1} \right)} \over 2}\,\,\,\,\,\,\,\,\left( {{\rm{arithmetic}}\,\,{\rm{sequence}}} \right)\) \({S_7} = 7 \cdot 4 = 28\,\,\, \Rightarrow \,\,\,\left\{ \matrix{ \,27 = 28  1 = \left( {1 + 2 + \ldots + 7} \right)  1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {N,M} \right) = \left( {7,2} \right)\,\,\,{\rm{with}}\,\,N > 2M\,\,\, \Rightarrow \,\,\,\,{\rm{viable!}} \hfill \cr \,25 = 28  3 = \left( {1 + 2 + \ldots + 7} \right)  \left( {1 + 2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {N,M} \right) = \left( {7,3} \right)\,\,\,{\rm{with}}\,\,N > 2M\,\,\, \Rightarrow \,\,\,\,{\rm{viable!}} \hfill \cr} \right.\) From the alternative choices given, we are sure we have found (EASILY!) the right answer! The correct answer is (D). We follow the notations and rationale taught in the GMATH method. Regards, Fabio. POSTMORTEM:Question 1.: how can we be sure that 26 (II) cannot be obtained? Please note that: \(\left. \matrix{ {S_8} = 4 \cdot 9 = 36 \hfill \cr {S_4} = 2 \cdot 10 = 10\,\,\, \hfill \cr} \right\}\,\,\, \Rightarrow \,\,\,26 = 36  10 = \left( {1 + 2 + \ldots + 7 + 8} \right)  \left( {1 + 2 + 3 + 4} \right)\) and the ONLY reason (N,M) = (8,5) must be refuted is the fact that N>2M is false... What about other values for (N,M)? How can we be SURE there is not a single viable possibility? Question 2.: which mathematical (GMATfocused) properties could be useful to find all possible (N,M) pairs for a given LARGER value? We will address both questions in our very next problem, here: https://gmatclub.com/forum/anumberis ... l#p2229235
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If M, N are integers greater than 1 such that 2M<N, which of the foll
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